Homework Help: Logarithmic spiral

1. Jul 26, 2012

martine

1. The problem statement, all variables and given/known data

a logarithmic spiral is given. The center lies on the x-axis. after a turn of 180 degrees counterclockwise I am 86.23m away from the starting point, after 360 degrees I'm 75.41m away from the start (radius, not along the spiral). Where am I after I walked exactly 3km along the spiral trajectory?

2. Relevant equations

equation: r=ae^b*theta
starting point corresponds to theta=0

3. The attempt at a solution

Um.. that's my problem. I never really learnt how to solve logarithmic calculations. Btw, this is not a homework, but part of a puzzle. I got the numbers from previous steps and found the equation online.

I understand that r=distance from origin, theta = angle with x-axis, a and b some constants. Right. But I don't see any way of working with this this equation or any of the expressions connected to it I could need a little push in the right direction.

Thanks a lot,
Martine

2. Jul 26, 2012

uart

Why don't you just write the two equations (with the known values of "r,theta" you have) and divide one equation by the other to eliminate "a" and solve for "b".

Hint1: "b" is negative in this case.

After you solve for the parameters "a" and "b" you then have to work out how to find the length of a path along the spiral.

Hint2: $dl = \left( \sqrt{r^2 + \left( dr/d\theta \right)^2} \right) d\theta$

3. Jul 26, 2012

HallsofIvy

You say "180 degrees" and "360 degrees" but radian measure should be use here.
when $\theta= \pi$, r= 86.23 so $86.23= ae^{b\pi}$. When $\theta= 2\pi$, r= 75.42 so $75.41= ae^{2b\pi}$. As uart says, dividing one equation by the other will cancel the 'a's leaving
$$\frac{75.41}{86.23}= \frac{e^{2b\pi}}{e^{b\pi}}= e^{b\pi}$$
that is easy to solve for b. Then use that value of b with either of the first equations to find a.

For the final problem you will need to integrate the "differential of arc length" that uart gave from 0 to $\theta$ and set it equal to 3000 to solve for $\theta$.